Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What's the derivative of the integral $$\int_1^x\sin(t) dt$$

Any ideas? I'm getting a little confused.

share|cite|improve this question

You can use the fundamental theorem of calculus, but if you have not yet covered that theorem, in short, you'll be taking the derivative - with respect to x - of the integral of $\sin(t)dt$ when the integral is evaluated from $1$ to $x$:

$$\frac{d}{dx}\left(\int_1^x \sin(t) \text{d}t\right) = \frac{d}{dx} [-\cos t]_1^x = \frac{d}{dx}\left(-\cos(x) - (-\cos(1))\right) = \sin(x).$$

and you'll no doubt be encountering the Fundamental Theorem of Calculus very, very soon:

For any integrable function $f$, and constant $a$: $$\frac{d}{dx} \left(\int_a^x f(t)dt \right)= f(x),$$

(provided $f$ is continuous at $x$).

share|cite|improve this answer
I think I've got it. d/dt(-cos(x)-(-cos(1)) = d/dt(-cos(x)+0) = sin(x) – Ian Nov 21 '12 at 19:35
Yes, you got it. – amWhy Nov 21 '12 at 19:36
Thanks for the help. – Ian Nov 21 '12 at 19:38
Your very welcome. – amWhy Nov 21 '12 at 19:41
Just be careful, $\cos(1) \neq 0$. – Emily Nov 21 '12 at 19:41

Using the fundamental theorem of calculus we know that the answer is $\sin(x)$

share|cite|improve this answer

$ \frac{d}{dt}\int_1^x\sin(t)dt = \frac{d}{dt} [-\cos t]_1^x = \frac{d}{dt}[-\cos x+\cos(1)] = \sin x $

share|cite|improve this answer

If $f$ is any function at all that can be integrated, then the derivative of the integral of $f(t)dt$ from $1$ to $x$ is $f(x)$. This wonderful fact is the Fundamental Theorem of Calculus.

share|cite|improve this answer
I'm not so sure. For a trivial counter-example, if $\frac{d}{dx} \int_a^x f(t) dt = f(x)$ for every $x \in [a,b]$, then we can modify f on a set of measure 0 (say some countable set) and obtain a function for which this does not hold. The Fundamental theorem of Calculus holds in general only if f is continuous. – anonymous Nov 23 '12 at 1:39

You can use a nice theorem called the Fundamental Theorem of Calculus . Here, we're mainly worried about FTC part 1. Below is a summary of what FTC part 1 says.

Let $f$ be a continuous, function defined on $[a,b]$ and $$F(x) := \int_a^x{f(t)} \ dt \quad \quad \forall x \in [a,b] $$

Then, $F$ is continuous on the closed interval and differentiable on the open interval $(a,b)$ and $F'(x) = f(x) \ \forall x \in (a,b)$.

So, for your problem:

$$\frac{d}{dx}\int_1^x\sin(t) \ dt = \sin(x) \cdot \frac{d}{dx} (x) = \sin(x)$$

Note that you have to replace the $t's$ in the integrand with an $x$ and multiply by the derivative of the upper bound, assuming your lower bound is constant (which it is here.)

As a demonstration of this, suppose we want to calculate

$$\frac{d}{dx}\int_1^{x^2}\sin(t) \ dt $$

This is the same as $$\sin(x) \cdot \frac{d}{dx}(x^2) = 2x \sin(x)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.